Question

an object is taken from a freezer at - 10°c. let t be the time in hours after the object was taken from the freezer. at time t the average temperature of the object is increasing at the rate of t(t)=20e^{-0.2t} degrees celsius per hour. find the temperature of the object at time t. t(t)=□

Explanation:

Step1: Integrate the rate function

We know that \(T(t)=\int T^{\prime}(t)dt\). Given \(T^{\prime}(t) = 20e^{- 0.2t}\), then \(T(t)=\int20e^{-0.2t}dt\).
Using the formula \(\int ae^{bt}dt=\frac{a}{b}e^{bt}+C\) (\(a = 20\), \(b=-0.2\)), we have \(T(t)=\frac{20}{-0.2}e^{-0.2t}+C=-100e^{-0.2t}+C\).

Step2: Use the initial condition

When \(t = 0\), \(T(0)=-10\). Substitute \(t = 0\) into \(T(t)=-100e^{-0.2t}+C\).
We get \(-10=-100e^{0}+C\). Since \(e^{0}=1\), then \(-10=-100 + C\), so \(C = 90\).

Answer:

\(T(t)=90 - 100e^{-0.2t}\)